= Solution
Use the <descending chain condition> on the finite intersections of <maximal right ideals>. It gives a minimal such intersection $K=M_1\cap\cdots\cap M_r$. Intersecting $K$ with any other <maximal right ideal> cannot decrease it, by minimality. Thus $K$ lies in every <maximal right ideal>, and $K=J(R)$. When $J(R)=0$, the diagonal map
$$
R_R\longrightarrow\bigoplus_{i=1}^rR/M_i,\qquad a\longmapsto(a+M_i)_i
$$
is injective, and each summand on the right is a <simple module>.
For completeness, a submodule $N$ of a finite direct sum of <simple modules> is semisimple. Induct on the number of summands. For $S\oplus T$ with $S$ simple, $N\cap S$ is either $0$ or $S$. In the first case projection into $T$ identifies $N$ with a submodule of $T$, so induction applies. In the second case subtracting the $S$ coordinate shows $N=S\oplus(N\cap T)$, and induction again applies. This also shows finite <module length>. Apply it to the image of the diagonal map to obtain
$$
\boxed{J(R)=0\text{ and }R\text{ right Artinian}\Longrightarrow R_R\text{ semisimple}.}
$$
This is the <semisimplicity of a right Artinian ring with zero radical>. It does not assume right Noetherianity in order to prove it.
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